MA 3416: Group representations
Selected answers/solutions to the assignment due March 4, 2015
1. Clearly, triv ⊗ W ' W for all W. Also, sgn ⊗ sgn ' triv, sgn ⊗ U ' U, sgn ⊗ V ' V 0 ,
sgn ⊗ V 0 ' V, as it is obvious from examining characters of these. It remains to compute
U ⊗ U, U ⊗ V, V ⊗ V, and V ⊗ V 0 . Note that U ⊗ V 0 ' U ⊗ sgn ⊗ V ' U ⊗ V, and
V 0 ⊗V 0 ' V ⊗sgn⊗V ⊗sgn ' V ⊗V because of the product formulas for the sign representation
above. Computing the scalar products of characters, we find U ⊗ U ' U ⊕ triv ⊕ sgn,
U ⊗ V ' V ⊕ V 0 , V ⊗ V ' V ⊕ V 0 ⊕ triv ⊕ U, V ⊗ V 0 ' V ⊗ V ⊗ sgn ' V ⊕ V 0 ⊕ sgn ⊕ U.
2. (a) The character is computed by direct inspection: we must figure out, for each type
of rotation, how many vertices that rotation keeps intact. The result is written below under
the character table of S4 .
triv
sign
V
V0
U
vertices
e
1
1
3
3
2
8
(ij)
1
-1
1
-1
0
0
(ijk)
1
1
0
0
-1
2
(ijkl)
1
-1
-1
1
0
0
(ij)(kl)
1
1
-1
-1
2
0
Computing the scalar products with irreducible characters, we see that this representation is
isomorphic to triv ⊕ sign ⊕ V ⊕ V 0 .
Finally, note that the set of all vertices of the cube is the union of the sets of vertices
of two regular tetrahedra, each formed by four vertices of the cube that have no edges of
the cube between them. The invariant subspaces are: constant functions, functions which
assume the same value on one of the tetrahedra and the same opposite value on the other,
even functions which add up to zero on each of the tetrahedra, odd functions that add up to
zero on each of the tetrahedra.
(b) Since T is an intertwining operator, and since all irreducible constituents of our representation are different, T acts by a scalar on each irreducible constituent, and we should
just compute each of these scalars to examine the dynamics of T . Clearly, on the constant
functions T acts as multiplication by 1, on functions that assume the same value on one of
the tetrahedra and the same opposite value on the other T acts as multiplication by −1 (each
a gets replaced by (−a − a − a)/3 = −a, on even functions that add up to zero on each of the
tetrahedra T acts as multiplication by −1/3 (each a gets replaced by (b + c + d)/3 = (−a)/3),
and on odd functions that add up to zero on each of the tetrahedra T acts as multiplication by
1/3 (each a gets replaced by (b + c + d)/3 = (−(−a))/3). Therefore, the limiting behaviour
of T applied to a certain vector is determined by the projections of that vector on the trivial
representation and the sign representation. Note that for the configuration in question the
sums of values on both tetrahedra are the same: 1 + 3 + 6 + 8 = 18 = 2 + 4 + 5 + 7. Therefore,
the projection on the sign representation, where the opposite values are proportional to the
difference of values on the two tetrahedra, is equal to zero, and the projection on the trivial
representation, the space of constant functions, is the function whose value on each vertex is
equal to 4.5. That projection is the limit lim T n (w).
n→∞
3. Note that if e1 , . . . , en is a basis of V, then the wedge products ei1 ∧ · · · ∧ eik with
i1 < · · · < ik form a basis of Λk (V). Assume that A can be diagonalised, and that f1 , . . . , fn
is a basis of V consisting of eigenvectors of A with eigenvalues λ1 , . . . , λn . Then the wedge
products fi1 ∧ · · · ∧ fik with i1 < · · · < ik are eigenvectors of Λk (A) with the respective
eigenvalues being λi1 λi2 · · · λik , and the trace of Λk (A) is equal to the sum ofQall of these.
Note that the characteristic polynomial of A, that is det(A − tI), is equal to ni=1 (λi − t).
Therefore, the trace of Λk (A) is equal, up to a sign (−1)n−k , to the coefficient of tn−k in that
polynomial.
4. Note that for a finite group G, the transformation ρ(g) can be diagonalised. Assume
that it has eigenvalues λ1 , . . . , λn . Then ρ(g2 ) = ρ(g)2 has eigenvalues λ21 , . . . , λ2n . Therefore,
X
1
1
(χV (g)2 − χV (g2 )) = ((λ1 + · · · + λn )2 − (λ21 + · · · + λ2n )) =
λi λj ,
2
2
i<j
which is precisely χΛ2 (V) (g), as we know from the previous question.
5. The characters of the corresponding representations are written below the character
table of S4 . Note that U is two-dimensional, so Λ2 (U) is one-dimensional, and the corresponding character is given by the determinant of the corresponding matrices acting on U,
and Λ3 (U) = 0. Similarly, V and V 0 are three-dimensional, so the exterior square Λ2 (V)
can be handled by the previous question, the exterior square Λ2 (V 0 ) is isomorphic to the
exterior square Λ2 (V), because we multiply by the sign twice, the action on the exterior cube
Λ3 (V) is given by the determinant, and the action on the exterior cube Λ3 (V 0 ) is given by
the determinant multiplied by the sign.
triv
sign
V
V0
U
2
Λ (V)
Λ2 (V 0 )
Λ2 (U)
Λ3 (V)
Λ3 (V 0 )
Λ3 (U)
e
1
1
3
3
2
3
3
1
1
1
0
(ij)
1
-1
1
-1
0
-1
-1
-1
-1
1
0
(ijk)
1
1
0
0
-1
0
0
1
1
1
0
(ijkl)
1
-1
-1
1
0
1
1
-1
-1
1
0
(ij)(kl)
1
1
-1
-1
2
-1
-1
1
1
1
0
Thus, Λ2 (V) ' Λ2 (V 0 ) ' V 0 , Λ2 (U) ' Λ3 (V) ' sign, and Λ3 (V 0 ) ' triv.